Question

Two gases, \(\mathrm{HBr}\) and \(\mathrm{CH}_{4}\), have molecular weights 81 and 16 , respectively. The HBr effuses through a certain small opening at the rate of \(4 \mathrm{ml} / \mathrm{sec}\). At what rate will the \(\mathrm{CH}_{4}\) effuse through the same opening?

Short answer

The rate at which \(\mathrm{CH}_{4}\) effuses through the same opening is \(9\, ml/sec\).

Step-by-step solution

Write down Graham's Law of Effusion

To solve this problem, we need to use Graham's law of effusion, which states that the rate of effusion of two gases is inversely proportional to the square root of their molar masses: $$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$$ Where: - \(r_1\) is the effusion rate of gas 1 (in this case, \(\mathrm{HBr}\)), - \(r_2\) is the effusion rate of gas 2 (in this case, \(\mathrm{CH}_{4}\)), - \(M_1\) is the molar mass of gas 1, - \(M_2\) is the molar mass of gas 2.

Plug in the given values

We are given the following values: - \(r_1 = 4\, ml/sec\) (the rate at which HBr effuses), - \(M_1 = 81\, g/mol\) (the molar mass of HBr), - \(M_2 = 16\, g/mol\) (the molar mass of CH4). We want to find \(r_2\), which is the rate at which \(\mathrm{CH}_{4}\) effuses. Substitute these values into Graham's law of effusion equation: $$\frac{4\, ml/sec}{r_2} = \sqrt{\frac{16\, g/mol}{81\, g/mol}}$$

Solve for the effusion rate of CH4

Now we need to solve the equation for \(r_2\). First, let's simplify the square root part: $$\sqrt{\frac{16}{81}} = \frac{4}{9}$$ Now, we have: $$\frac{4\, ml/sec}{r_2} = \frac{4}{9}$$ To solve for \(r_2\), we can multiply both sides by \(r_2\) and by 9: $$4\, ml/sec \times 9 = 4r_2$$ Now, divide both sides by 4: $$9\, ml/sec = r_2$$

State the final answer

The rate at which \(\mathrm{CH}_{4}\) effuses through the same opening is \(9\, ml/sec\).

Key concepts

Molar Mass

The molar mass is a fundamental concept in chemistry that represents the mass of one mole of a given substance. It is generally expressed in grams per mole (g/mol). In simpler terms, a mole is a way to count particles, similarly to a dozen counting eggs. Knowing the molar mass allows us to relate the mass of a substance to the number of particles contained in that mass. Molar mass is calculated by adding up the atomic masses of all the atoms in a molecule, which can be found on the periodic table. For example, the molar mass of hydrogen bromide (\[\mathrm{HBr}\]) is calculated by summing the atomic mass of hydrogen (approximately 1 g/mol) and bromine (approximately 80 g/mol), giving a total molar mass of 81 g/mol as indicated in the original exercise. Understanding molar mass is crucial for solving problems in stoichiometry, and it is an important parameter in Graham's Law of Effusion, which compares the effusion rates of different gases based on their molar masses.

Molecular Weight

Molecular weight is often used interchangeably with molar mass, as they both refer to the sum of the atomic weights of atoms in a molecule, measured in atomic mass units (amu) or grams per mole. However, there is a subtle difference between them. While molecular weight refers strictly to the weight of a molecule at the atomic scale, molar mass refers to the mass of a mole of particles, making it a more practical unit for laboratory calculations. For instance, in our exercise, the molecular weight of methane (\(\mathrm{CH_{4}}\)) is calculated by adding the atomic mass of one carbon atom (approximately 12 amu) and four hydrogen atoms (approximately 1 amu each), leading to a molecular weight of 16 amu. This value is directly comparable to the molar mass when considering bulk amounts like those involved in effusion processes.Understanding molecular weight is key to working with chemical equations and conversions. It assists in predicting how gases behave under various conditions, especially in calculations involving Graham's Law of Effusion, where these values determine the rates at which gases pass through tiny openings.

Effusion Rate

Effusion is the process by which gas molecules escape through a small opening. The effusion rate, therefore, is the speed at which this process occurs. According to Graham's Law of Effusion, which is crucial in this context, the effusion rate of a gas is inversely proportional to the square root of its molar mass.This means that lighter gases effuse more quickly than heavier gases. Hence, observing the equation:\[\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \]You can see that knowing the molar masses of two gases allows us to solve for their relative effusion rates. In the given example, by using Graham's Law, we calculated that the effusion rate of methane would be faster than that of hydrogen bromide, with methane effusing at 9 ml/sec compared to hydrogen bromide's 4 ml/sec. Understanding effusion rates is vital for applications in chemistry and industry, helping to design systems for separating gases and analyzing gas mixtures.

Gases

Gases are a state of matter consisting of particles that have neither a defined shape nor a defined volume. They fill any container they are placed in, as their particles move freely and are in constant motion. This property is particularly important when discussing concepts like effusion, as it explains how gases spread and move through openings.Unlike solids or liquids, the behavior of gases is influentially impacted by conditions like temperature and pressure, which are described by various gas laws, including the ideal gas law. These principles explain how gas particles interact with their environments and with each other. Graham's Law of Effusion, which we’ve been focusing on, specifically looks at gases and their molecular differences. In this exercise, we are comparing two different gases, hydrogen bromide (\(\mathrm{HBr}\)) and methane (\(\mathrm{CH_4}\)), to understand how their composition affects their movement through an opening. Understanding the behavior of gases is fundamental in many scientific and industrial applications, from the design of pneumatic systems to the analysis of atmospheric processes.